I am looking for some help with a motorbike brake light system. I have build some led arrays, 6 leds each in series, then in parallel for a total of 48 led's per board, both boards wired in series.

A picture says a thousand words so here is the "circuit" Diagram:

Sorry for my crappy drawing, it was getting late and I was frustrated heh.

The reason for the resistors is that when the headlights are turned on the "tail" line is energised, at a lower current so that the LED arrays illumante dimmer, then when the brake is pressed, the "brake" line is energised, bypassing the resistors and getting brighter. This works when one array is connected, but is back feeding with more than one connected.

So when I connect the second LED strip (at the bottom of the diag), the LED arrays go out (when on tail mode) when on braking mode everything lights up as it should. Its my thinking that I need some diodes at A and B on the diagram, can anybody confirm if I am on the right track

Additionally when connecting the second led strip, when turning on the tail light, the brake lights up as well, I believe due to the lack of the diode, the 13v tail line is energising the brake line (albeit with reduced current because of the resistor)

First, please repost your diagram so that it shows the entire circuit.

Second, independant of your need for diodes, you are not taking into account the minimum voltage that the LEDs would see - for example, what if your battery has a shorted cell, and the motoris stopped. You might have less than 10 Volts under these conditions,and the LEDs might not light at all, and if they do, with reduced brightness.Car Manufacturers use LED circuits that are designed so that they will work even under low voltage conditions - 8 Volts or less, and use what is called a constant current circuit so that the brightness of the LEDs is independant of the battery voltage.

2) The constant current circuit will supply the same current to the diode string regardless of what the supply voltage is, as long as it is above the sum of the voltage across the diodes and what the constant current circuit needs - a simple one is an NPN transistorwith a resistor in it's emitter lead to ground, and a string of 3 LEDs connected to the transistor collector and +12V. If the transistorhas a turn on voltage of say .7V, and you have a 10 ohm resistorin the emitter lead, when you supply say .8V between thebase of the transistor and ground, the transistor will drop .7Vand there will be .1V dropped arcoss the emitter resistor. Ohms law tells us that .1V across 10 ohms gives 10mA,so that 10mA flows thru the LEDs. It dosn't matter now if the 12V changes to 8V or 20V, only 10mA flows thru the LEDs.

Change the voltage applied to the transistor's base to 1V, and 30mAwill flow thru the emitter resistor [and the LEDs].

Change the emitter resistor to 1 ohm, and either 100 or 300mA flowsthru the resistor [and LEDs]. connect a 470 ohm resistor to the batteryand connect two 1n4004 diodes in series to ground [cathodes towards ground],and the juction of the 470 ohm resistor and the 'upper 1n4004 diode tothe base of the transistor and you have a 1.2 - 1.4V voltage referenceto drive the transistor. That gives you .5V to .7V across the emitterresistor for 50/70mA or 500/700mA thru the LEDs. [note that youneed a power transistor that has a heatsink attached, because the transistor has [50/70mA or 500/700mA times the voltage acrossthe transistor] in watts being disipated by the transistor.[The higher the battery voltage, the more power is disipated by the transistor also].

Don't forget to calculate the wattages the resistors are disipating.

Draw the circuit or google 'emitter follower' and think about it.

There are alternative circuits, but that is one way to build a constantcurrent circuit.